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Let's keep building our table of Laplace transforms. And now

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we'll do a fairly hairy problem, so I'm going to have

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to focus so that I don't make a careless mistake.

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But let's say we want to take the Laplace transform-- and

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this is a useful one.

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Actually, all of them we've done so far are useful.

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I'll tell you when we start doing not-so-useful ones.

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Let's say we want to take the Laplace transform of the sine

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of some constant times t.

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Well, our definition of the Laplace transform, that says

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that it's the improper integral.

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And remember, the Laplace transform is just a

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definition.

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It's just a tool that has turned out to

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be extremely useful.

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And we'll do more on that intuition later on.

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But anyway, it's the integral from 0 to infinity of e to the

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minus st, times-- whatever we're taking the Laplace

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transform of-- times sine of at, dt.

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And now, we have to go back and find our integration by

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parts neuron.

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And mine always disappears, so we have to reprove

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integration by parts.

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I don't recommend you do this all the time.

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If you have to do this on an exam, you might want to

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memorize it before the exam.

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But always remember, integration by parts is just

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the product rule in reverse.

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So I'll just do that in this corner.

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So the product rule tells us if we have two

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functions, u times v.

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And if I were take the derivative of u times v.

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Let's say that they're functions of t.

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These are both functions of t.

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I could have written u of x times u of x.

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Then that equals the derivative of the first times

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the second function, plus the first function times the

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derivative of the second.

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Now, if I were to integrate both sides, I get uv-- this

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should be review-- is equal to the integral of u prime v,

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with respect to dt-- but I'm just doing a little bit of

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shorthand now-- plus the integral of uv prime.

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I'm just trying to help myself remember this thing.

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And let's take this and subtract it from both sides.

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So we have this integral of u prime v is going to be equal

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to this, uv minus the integral of uv prime.

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And, of course, this is a function of t.

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There's a dt here and all of that.

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But I just have to do this in the corner of my page a lot,

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because I always forget this, and with the primes and the

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integrals and all that, I always forget it.

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One way, if you did want to memorize it, you said, OK, the

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integration by parts says if I take the integral of the

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derivative of one thing and then just a regular function

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of another, it equals the two functions times each other,

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minus the integral of the reverse.

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Right?

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Here, when you take the subtraction, you're taking the

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one that had a derivative, now it doesn't.

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And the one that didn't have a derivative, now it does.

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But anyway, let's apply that to our problem at

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hand, to this one.

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Well, we could go either way about it.

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Let's make u prime is equal to-- we'll do our definition--

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u prime is equal to e to the minus st, in which case you

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would be the antiderivative of that, which is equal to minus

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1 over s e to the minus st, right?

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And actually, this is going to be an integration by parts

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twice problem, so I'm just actually going to define the

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Laplace transform as y.

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That'll come in useful later on.

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And I think I actually did a very similar example to this

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when we did integration by parts.

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But anyway, back to the integration by parts.

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So that's u.

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And let me do v in a different color.

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So when v-- if this is u prime, right?

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This is u prime, then this is v.

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So v is equal to sine of at.

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And then what is v prime?

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Well, that's just a cosine of at, right?

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The chain rule.

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And now, we're ready to do our integration.

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So the Laplace transform, and I'll just say that's y, y is

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equal to-- y is what we're trying to solve for, the

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Laplace transform of sine of at-- that is

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equal to u prime v.

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I defined u prime in v, right?

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That's equal to that.

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The integral of u prime times v.

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That equals uv.

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So that's minus 1 over s e to the minus st, times v, sine of

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at, minus the integral.

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And when you do the integration by parts, this

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could be an indefinite integral, an improper

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integral, a definite integral, whatever.

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But the boundary stays.

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And we can still say, from 0 to infinity of uv prime.

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So u is minus 1 over s e to the minus st, times v prime,

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times a cosine of at-- fair enough-- dt.

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Well, now we have another hairy

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integral we need to solve.

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So this might involve another integration by

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parts, and it does.

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Let's see if we can simplify it at [? all. ?]

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Let's take the constants out first. Let me

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just rewrite this.

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So we get y is equal to minus e to the minus st

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over s, sine of at.

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So you have a minus minus plus a over s-- a divided by s, and

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then these two negative signs cancel out-- times the

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integral from 0 to infinity, e to the minus st,

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cosine of at, dt.

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Let's do another integration by parts.

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And I'll do this in a purple color, just so you know this

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is our second integration by parts.

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Over here.

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Let's define once again, u prime is equal to e the minus

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st. So this is u prime.

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Then u is equal to minus 1 over s e to the minus st.

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We'll make v equal to cosine of at.

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The hardest part about this is not making careless mistakes.

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And then v prime-- I just want it to be on the same row-- is

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equal to minus a sine of at, right?

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The chain rule, derivative of cosine is minus sign.

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So let's substitute that back in, and we get-- this is going

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to get hairy; actually, it already is hairy-- y is equal

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to minus e to the minus st over s, sine of at, plus a

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over s, times-- OK.

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Integration by parts.

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uv.

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So that's minus 1 over s e to the minus st, times v, times

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cosine at, minus the integral from 0 to infinity.

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This problem is making me hungry.

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It's taking so much glucose from my bloodstream.

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I'm focusing so much not to make careless mistakes.

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Anyway, integral from 0 to infinity.

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And now, we have uv prime, so u is minus 1 over s e to the

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minus st. That's u.

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And then v prime times minus a.

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So let's make that minus cancel out with this one.

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So that becomes a plus.

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a sine of at, dt.

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I'm starting to see the light at the end of the tunnel.

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So then, let's simplify this thing.

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And, of course, we're going to have to evaluate this whole

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thing, right?

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Actually, we're going to have to evaluate everything.

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Let's just focus on the indefinite integral for now.

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We're going to have to take this whole thing and

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evaluate-- let's just say that y is the antiderivative and

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then evaluate it from infinity to 0.

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From 0 to infinity.

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So y is equal to minus e to the minus st

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over s, sine of at.

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Now let's distribute this.

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Minus a over s squared, e to the minus st, cosine of at.

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Right?

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OK, now I want to make sure I don't make a careless mistake.

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OK.

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Now, let's multiply this times this and take all of the

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constants out.

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So we have an a and an s.

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a over s.

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There's a minus sign.

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We have a plus a to the s.

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So we'll have a minus a squared over s squared, times

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the integral from 0-- well, I said I'm just worrying about

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the indefinite integral right now, and we'll evaluate the

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boundaries later.

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e to the minus st, sine of at, dt.

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Now, this is the part, and we've done this before, it's a

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little bit of a trick with integration by parts.

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But this expression, notice, is the same thing as our

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original y.

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Right?

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This is our original y.

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And we're assuming we're doing the indefinite integral, and

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we'll evaluate the boundaries later.

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Although we could have kept the boundaries the whole time,

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but it would have made it even hairier.

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So we can rewrite this integral as y.

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That was our definition.

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And actually, I just realized I'm running out of time, so

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I'll continue this hairy problem in the next video.

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See you soon.

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